- Largest remainder method
The largest remainder method is one way of allocating seats proportionally for representative assemblies with party list
voting systems . It is a contrast to thehighest averages method .Quotas
There are several possibilities for the quota. The most common are: the
Hare quota and theDroop quota .The Hare (or simple) Quota is defined as follows
:
The Hamilton method of apportionment is actually a largest-remainder method which is specifically defined as using the Hare Quota, named after
Alexander Hamilton , who invented the largest-remainder method, in1792 . It is used for legislative elections inRussia (with 7% exclusion threshold since 2007),Ukraine (3% threshold),Namibia and in the territory of Hong Kong. It was historically applied for congressional apportionment in theUnited States during thenineteenth century .The
Droop quota is the integer part of :and is applied in elections in South Africa. TheHagenbach-Bischoff quota is similar, being :either used as a fraction or rounded up.The Hare quota tends to be slightly more generous to less popular parties and the Droop quota to more popular parties, in the extent in which it is arguably considered more proportional than Droop quota [http://www.parl.gc.ca/information/library/PRBpubs/bp334-e.htm] [http://polmeth.wustl.edu/polanalysis/vol/8/PA84-381-388.pdf] [http://www.dur.ac.uk/john.ashworth/EPCS/Papers/Suojanen.pdf] [http://users.ox.ac.uk/~sann2300/041102-ceg-electoral-consequences-lijphart.shtml] [http://janda.org/c24/Readings/Lijphart/Lijphart.html] although it is more likely to give fewer than half the seats to a list with more than half the vote.
The
Imperiali quota :is rarely used since it suffers from the problem that it may result in more candidates being elected than there are seats available (this can also occur with theHagenbach-Bischoff quota but it's very unlikely and is impossible with the Hare and Droop quotas). This will certainly happen if there are only two parties. In such a case, it is usual to increase the quota until the number of candidates elected is equal to the number of seats available, in effect changing the voting system to the Jefferson apportionment formula (seeD'Hondt method ).Examples
These examples take an election to allocate 10 seats where there are 100,000 votes.
Hare quota
Party Yellows Whites Reds Greens Blues Pinks Total Votes 47,000 16,000 15,800 12,000 6,100 3,100 100,000 Seats 10 Hare Quota 10,000 Votes/Quota 4.70 1.60 1.58 1.20 0.61 0.31 Automatic seats 4 1 1 1 0 0 7 Remainder 0.70 0.60 0.58 0.20 0.61 0.31 Highest Remainder Seats 1 1 0 0 1 0 3 Total Seats 5 2 1 1 1 0 10 Droop quota
Party Yellows Whites Reds Greens Blues Pinks Total Votes 47,000 16,000 15,800 12,000 6,100 3,100 100,000 Seats 10 Droop Quota 9,091 Votes/Quota 5.170 1.760 1.738 1.320 0.671 0.341 Automatic seats 5 1 1 1 0 0 8 Remainder 0.170 0.760 0.738 0.320 0.671 0.341 Highest Remainder Seats 0 1 1 0 0 0 2 Total Seats 5 2 2 1 0 0 10 Pros and cons
It is very easy for the average voter to understand how Largest Remainder allocates seats. Provided the Hare quota is used, it gives no advantage to lists with either a large or a small proportion of the votes - to that extent it is neutral. However, whether a list gets an extra seat or not is highly dependent on how the votes are distributed among other parties; it is quite possible for a party to make a slight percentage gain yet lose a seat. A related paradox is that increasing the number of seats may cause a party to lose a seat (the so-called
Alabama paradox ). TheSainte-Laguë method avoids these paradoxes but is less easy for the average voter to understand.Technical evaluation and paradoxes
The largest remainder method is the only apportionment that satisfies the
quota rule ; in fact, it is designed to satisfy this criterion. However, it comes at the cost of paradoxical behaviour. TheAlabama paradox is defined as when an increase in seats apportioned leads to decrease in the number of seats a certain party holds. Suppose we want to apportion 25 seats between 6 parties in the proportions 1500:1500:900:500:500:200. The two parties with 500 votes get three seats each. Now allocate 26 seats, and it will be found that the these parties get only two seats apiece.With 25 seats, we get:
Party A B C D E F Total Votes 1500 1500 900 500 500 200 5100 Seats 25 Hare Quota 204 Quotas Received 7.35 7.35 4.41 2.45 2.45 0.98 Automatic seats 7 7 4 2 2 0 22 Remainder 0.35 0.35 0.41 0.45 0.45 0.98 Surplus seats 0 0 0 1 1 1 3 Total Seats 7 7 4 3 3 1 25 With 26 seats, we have:
Party A B C D E F Total Votes 1500 1500 900 500 500 200 5100 Seats 26 Hare Quota 196 Quotas Received 7.65 7.65 4.59 2.55 2.55 1.02 Automatic seats 7 7 4 2 2 1 23 Remainder 0.65 0.65 0.59 0.55 0.55 0.02 Surplus seats 1 1 1 0 0 0 3 Total Seats 8 8 5 2 2 1 26 ee also
*
List of democracy and elections-related topics External links
* [http://www.cut-the-knot.org/Curriculum/SocialScience/AHamilton.shtml Hamilton method experimentation applet] at
cut-the-knot
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